1. **State the problem:** We have the curve $$y = 2x + 8 - \frac{5}{x^2}$$ for $$x > 0$$. Points $$P$$ and $$Q$$ lie on this curve at $$x=1$$ and $$x=4$$ respectively. We want to find the exact area of the region $$R$$ bounded by the curve $$C$$ and the straight line joining $$P$$ and $$Q$$. 2. **Find coordinates of points $$P$$ and $$Q$$:** At $$x=1$$: $$y_P = 2(1) + 8 - \frac{5}{1^2} = 2 + 8 - 5 = 5$$ So, $$P = (1,5)$$. At $$x=4$$: $$y_Q = 2(4) + 8 - \frac{5}{4^2} = 8 + 8 - \frac{5}{16} = 16 - \frac{5}{16} = \frac{256}{16} - \frac{5}{16} = \frac{251}{16}$$ So, $$Q = \left(4, \frac{251}{16}\right)$$. 3. **Find equation of the straight line $$PQ$$:** Slope $$m = \frac{y_Q - y_P}{4 - 1} = \frac{\frac{251}{16} - 5}{3} = \frac{\frac{251}{16} - \frac{80}{16}}{3} = \frac{\frac{171}{16}}{3} = \frac{171}{48} = \frac{57}{16}$$. Equation using point-slope form at $$P$$: $$y - 5 = \frac{57}{16}(x - 1)$$ $$\Rightarrow y = 5 + \frac{57}{16}(x - 1) = 5 + \frac{57}{16}x - \frac{57}{16} = \frac{80}{16} + \frac{57}{16}x - \frac{57}{16} = \frac{57}{16}x + \frac{23}{16}$$ So, line $$PQ: y = \frac{57}{16}x + \frac{23}{16}$$. 4. **Set up the integral for the area $$R$$:** Area between curve and line from $$x=1$$ to $$x=4$$ is $$\text{Area} = \int_1^4 \left( \text{curve} - \text{line} \right) dx = \int_1^4 \left( 2x + 8 - \frac{5}{x^2} - \left( \frac{57}{16}x + \frac{23}{16} \right) \right) dx$$ Simplify the integrand: $$2x + 8 - \frac{5}{x^2} - \frac{57}{16}x - \frac{23}{16} = \left(2x - \frac{57}{16}x\right) + \left(8 - \frac{23}{16}\right) - \frac{5}{x^2} = \left(\frac{32}{16}x - \frac{57}{16}x\right) + \left(\frac{128}{16} - \frac{23}{16}\right) - \frac{5}{x^2} = -\frac{25}{16}x + \frac{105}{16} - \frac{5}{x^2}$$ 5. **Integrate term-by-term:** $$\int_1^4 \left(-\frac{25}{16}x + \frac{105}{16} - \frac{5}{x^2} \right) dx = \int_1^4 -\frac{25}{16}x \, dx + \int_1^4 \frac{105}{16} \, dx - \int_1^4 \frac{5}{x^2} \, dx$$ Calculate each integral: - $$\int_1^4 -\frac{25}{16}x \, dx = -\frac{25}{16} \cdot \frac{x^2}{2} \Big|_1^4 = -\frac{25}{32} (16 - 1) = -\frac{25}{32} \times 15 = -\frac{375}{32}$$ - $$\int_1^4 \frac{105}{16} \, dx = \frac{105}{16} (4 - 1) = \frac{105}{16} \times 3 = \frac{315}{16}$$ - $$\int_1^4 \frac{5}{x^2} \, dx = 5 \int_1^4 x^{-2} \, dx = 5 \left(-\frac{1}{x}\right) \Big|_1^4 = 5 \left(-\frac{1}{4} + 1\right) = 5 \times \frac{3}{4} = \frac{15}{4}$$ 6. **Sum the results:** $$\text{Area} = -\frac{375}{32} + \frac{315}{16} - \frac{15}{4}$$ Convert all to denominator 32: $$-\frac{375}{32} + \frac{315 \times 2}{32} - \frac{15 \times 8}{32} = -\frac{375}{32} + \frac{630}{32} - \frac{120}{32} = \frac{-375 + 630 - 120}{32} = \frac{135}{32}$$ **Final answer:** $$\boxed{\frac{135}{32}}$$ This is the exact area of the region $$R$$ bounded by the curve and the line segment joining points $$P$$ and $$Q$$.